Pretty conjecture $x^{\left(\frac{y}{x}\right)^n}+y^{\left(\frac{x}{y}\right)^n}\leq 1$

Solution 1:

I am sorry for not proving your conjecture, but I thought of writing up my thoughts as it might help you.

Without loss of generality, we can say that $y \geq 0.5$. Let $q=y/x$ and because $y \geq 0.5$ we have $q\geq 1$ (you can do everything in reverse with $x \geq 0.5$ and $q\leq 1$). Let $a_0=x$ and $b_0=y$. Then we can write: $$ x^{\left( \frac{y}{x} \right)^n} + y^{\left(\frac{x}{y}\right)^n}=x^{q^n}+y^{\left(\frac{1}{q}\right)^n}=a_n+b_n, $$ with $$\begin{align} a_n &= a_{n-1}^q,\\ b_n &= b_{n-1}^{1/q}. \end{align}$$

It is easy to verify that with $b_n\geq0.5$, also $b_{n+1}\geq 0.5$. Now, let us assume that $a_{n-1} \leq 1-b_{n-1}$. For $n=1$, we have $a_0 \leq 1-b_0$ (more specifically, $a_0=1-b_0$). Because of that, we can write (remember that $q \geq 1$): \begin{align} a_{n+1} + b_{n+1} &= a_n^q+b_n^{1/q} \\ &\leq (1-b_n)^q + b_n^{1/q}. \end{align}

What is left, is to prove that $$ (1-b_n)^q+b_n^{1/q} \leq 1, \quad \forall \,\,\, b_n \in [0.5, 1], q\geq1 $$

This is where I got stuck, but perhaps you know how to continue from here. In the figure below, I plotted $(1-b)^q+b^{1/q}$ with $b$ on the interval $[0.5,1]$ for various values of $q$ and it seems that the result is always equal or below $1$.

Solution 2:

Partial answer

Let us prove the case when $0 < x \le \frac{14}{33}$.

By Bernoulli inequality, we have $$y^{(x/y)^n} = (1 - x)^{(x/y)^n} \le 1 - x(x/y)^n.$$ It suffices to prove that $$x^{(y/x)^n} + 1 - x(x/y)^n \le 1$$ or $$\ln x \le \frac{- \ln[(y/x)^n]}{(y/x)^n - 1}.$$ Since $u \mapsto \frac{-\ln u}{u - 1}$ is strictly increasing on $u > 1$, it suffices to prove that $$\ln x \le \frac{- \ln[(y/x)]}{(y/x) - 1}.$$ With the substitution $x = \frac{1}{z}$, it suffices to prove that, for all $z \ge \frac{33}{14}$, $$z\ln(z) - \ln(z - 1) - 2\ln z \ge 0$$ which is true (easy).